James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

500 Chapter 7 Techniques of Integration
which has the solution A − 1, B − 21, C − 21, D − 1, and E − 0. Thus
y 1 2 x 1 2x 2 2 x 3
xsx 2 1 1d 2 dx − y S 1 x 2 x 1 1
x 2 1 1 1
x
sx 2 1 1d 2D dx
In the second and fourth terms we made
the mental substitution u − x 2 1 1.
− y dx
x 2 y
x
x 2 1 1 dx 2 y
dx
x 2 1 1 1 y
− ln | x | 2 1 2 lnsx 2 1 1d 2 tan 21 x 2
x dx
sx 2 1 1d 2
1
2sx 2 1 1d 1 K
n
NOTE Example 8 worked out rather nicely because the coefficient E turned out to
be 0. In general, we might get a term of the form 1ysx 2 1 1d 2 . One way to integrate such
a term is to make the substitution x − tan . Another method is to use the formula in
Exercise 72.
Sometimes partial fractions can be avoided when integrating a rational function. For
instance, although the integral
x
y
2 1 1
xsx 2 1 3d dx
could be evaluated by using the method of Case III, it’s much easier to observe that if
u − xsx 2 1 3d − x 3 1 3x, then du − s3x 2 1 3d dx and so
Rationalizing Substitutions
y
x 2 1 1
xsx 2 1 3d dx − 1 3 ln | x 3 1 3x | 1 C
Some nonrational functions can be changed into rational functions by means of appropriate
substitutions. In particular, when an integrand contains an expression of the form
s n tsxd, then the substitution u − s n tsxd may be effective. Other instances appear in the
exercises.
Example 9 Evaluate y sx 1 4
x
dx.
SOLUTION Let u − sx 1 4 . Then u 2 − x 1 4, so x − u 2 2 4 and dx − 2u du.
Therefore
y sx 1 4
x
dx − y
u
u 2 2 4 2u du − 2 y
u 2
u 2 2 4 du − 2 y S1 1 4 du
u 2 2 4D
We can evaluate this integral either by factoring u 2 2 4 as su 2 2dsu 1 2d and using
partial fractions or by using Formula 6 with a − 2:
y s x 1 4
x
dx − 2 y du 1 8 y
− 2u 1 8 ?
du
u 2 2 4
1
2 ? 2
Z ln u 2 2
u 1 2
Z 1 C
− 2sx 1 4 1 2 ln Z
sx 1 4 2 2
sx 1 4 1 2
Z 1 C
n
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